Question

Let $X$ be a random variable having probability density function $f_{X}(x)=\begin{cases}2x & 0<x<1 \\ 0 & \text{otherwise}\end{cases}$ then the density of $Y=\frac{1}{X^{\frac{1}{\alpha}}}$ is

• A. $f_{Y}(y)=\frac{\alpha}{y^{\alpha+1}}; y>1$
• B. $f_{Y}(y)=\frac{\alpha}{2y^{\frac{\alpha}{2}+1}}; y>1$
• C. $f_{Y}(y)=\frac{2\alpha}{y^{2\alpha+1}}; y>1$
• D. $f_{Y}(y)=\frac{1}{2\alpha y^{\frac{1}{2\alpha}+1}}; y>1$

Hint:

When performing a transformation $Y=g(X)$, the PDF of $Y$ is $f_Y(y) = f_X(x) |\frac{dx}{dy}|$. Always ensure you substitute $x$ entirely with its equivalent expression in $y$ and re-calculate the support (range) for the new variable.

Answer 1

The Correct Option is C

To find the probability density function of $Y$, we use the method of transformation of variables.

Step 1: \color{redDefine the Transformation and Inverse
Given $Y = \frac{1}{X^{1/\alpha}} = X^{-1/\alpha}$.
To find $X$ in terms of $Y$:
$Y^{-\alpha} = (X^{-1/\alpha})^{-\alpha} = X$.
So, the inverse transformation is $x = g^{-1}(y) = y^{-\alpha}$.

Step 2: \color{redDetermine the Range of Y
The original variable $X$ is defined on $(0, 1)$.
As $x \to 0^{+}$, $y = (0^{+})^{-1/\alpha} \to \infty$.
As $x \to 1^{-}$, $y = (1^{-})^{-1/\alpha} \to 1$.
Thus, $Y$ is defined for $y > 1$.

Step 3: \color{redCalculate the Jacobian of the Transformation
The Jacobian $|J|$ is $|\frac{dx}{dy}|$:
$\frac{dx}{dy} = \frac{d}{dy}(y^{-\alpha}) = -\alpha y^{-\alpha-1}$.
So, $|J| = |-\alpha y^{-\alpha-1}| = \alpha y^{-\alpha-1}$ (since $\alpha > 0$ and $y > 1$).

Step 4: \color{redApply the Transformation Formula
The density of $Y$ is $f_Y(y) = f_X(g^{-1}(y)) \cdot |J|$.
Substitute $x = y^{-\alpha}$ into $f_X(x) = 2x$:
$f_Y(y) = 2(y^{-\alpha}) \cdot (\alpha y^{-\alpha-1})$.
$f_Y(y) = 2\alpha y^{-\alpha - \alpha - 1} = 2\alpha y^{-2\alpha - 1}$.
$f_Y(y) = \frac{2\alpha}{y^{2\alpha+1}}$ for $y > 1$.
This matches Option (3).